finalsolns - APPM 2360 Final Exam May 9, 2002 1 On the...

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Unformatted text preview: APPM 2360 Final Exam May 9, 2002 1 On the front of your bluebook, write your name, instructor’s name, and recitation number. There are NINE QUESTIONS . You must WORK EIGHT OUT OF NINE problems. CLEARLY IDENTIFY THE EIGHT PROBLEMS YOU WANT GRADED. Start each problem on a new page, SHOW ALL YOUR WORK in your bluebook, and box in your answers. Calculators, books, and crib sheets are NOT permitted. 1. (25 points) Indicate whether the following statements are TRUE or FALSE . No justification is necessary. a. (5 points) The equation x = x 2- 1 has an equilibrium at x = 1 that is a source. b. (5 points) The initial-value problem y = y (1- y ) 1 / 3 with y ( t ) = y has a unique solution for all ( t ,y ). c. (5 points) The system of equations 1 5 15- 2- 2 8 1 4- 3- 9 2112 6 4 7 8- 8 x 1 x 2 x 3 x 4 = 11 8- 7- 13 has a unique solution. d. (5 points) The vector functions v 1 = e t e t , v 2 = e t are linearly dependent. e. (5 points) Consider the homogeneous 2 × 2 system with constant coefficients x = Ax . Sup- pose λ 1 and λ 2 are the eigenvalues of A and Re λ 1 = Re λ 2 =- 5, then the equilibrium point x = must be a spiral sink. SOLUTION a. True. b. False. c. False. d. False. e. False. APPM 2360 Final Exam May 9, 2002 2 2. (25 points) Solve the following differential equations: a. (10 points) y = 3 e 2 ln t + y b. (15 points) y + y = e- t sin t SOLUTION a. The ODE is separable: y = 3 t 2 e y ⇒ e- y y = 3 t 2 . Integrate with respect to t : Z e- y dy = Z 3 t 2 dt- e- y = t 3 + C 1 e- y =- t 3 + C 2 e y = 1 C 2- t 3 y ( t ) = ln 1 C 2- t 3 b. The ODE is linear. Thus, we can use either variation of parameters or an integrating factor. Variation of Parameters: Homogeneous solution : y h ( t ) = Ce- t Particular solution : y p ( t ) = C ( t ) e- t Substitute the particular solution into the ODE and obtain C = sin t C ( t ) =- cos t + K This gives particular solution y p ( t ) = (- cos t + K ) e- t . Ignoring or absorbing con- stant K gives the general solution y ( t ) = C 1 e- t- e- t cos t. Integrating Factor: Integrating factor : e t e t ( y + y ) = e t (sin te- t ) ( e t y ) = sin t e t y =- cos t + C y ( t ) = Ce- t- e- t cos t. APPM 2360 Final Exam May 9, 2002 3 3. (25 points) Solve the following initial-value problem: x 00 = 2 x 3 , x (0) = 2 , x (0) =- 4 . SOLUTION The ODE x 00 = 2 x 3 has the form x 00 = f ( x ). Thus, multiply the ODE by x and obtain x x 00 = x (2 x 3 ) , 1 2 x 2 = x (2 x 3 ) . Integrate with respect to t : 1 2 x 2 = Z (2 x 3 ) dx, 1 2 x 2 = 1 2 x 4 + C. Using the initial condition x (0) = 2 and x (0) =- 4, or equivalently, ( t ,x ,x ) = (0 , 2 ,- 4), we get 1 2 (- 4) 2 = 1 2 (2) 4 + C....
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This note was uploaded on 06/29/2008 for the course APPM 2360 taught by Professor Williamheuett during the Summer '07 term at Colorado.

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finalsolns - APPM 2360 Final Exam May 9, 2002 1 On the...

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